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Is k^2 +k -2 > 0 ?

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Is k^2 +k -2 > 0 ? [#permalink]

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Is k^2 + k - 2 > 0 ?

(1) k < 1
(2) k < -2
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Feb 2013, 15:57, edited 1 time in total.
Edited the question.
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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Is k^2 + k - 2 > 0 ?

First let's see for which ranges of k the given inequality holds true: \(k^2 + k - 2 > 0\) --> \((k+2)(k-1)>0\) --> roots are -2 and 1 --> ">" sign indicates that the solution lies to the left of a smaller root and to the right of a larger root (check here for this technique: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus the given inequality holds true for: \(k<-2\) and \(k>1\).

So, the question asks whether \(k<-2\) or \(k>1\).

(1) k < 1. Not sufficient.
(2) k < -2. Sufficient.

Answer: B.
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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New post 12 Feb 2013, 16:11
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rephrasing the statement we have

\((x + 2) (x - 1) > 0\) ====> \(x > -2\) AND \(x > 1\)

on a number line we have

-------------------0 ----- \(1\) +++++++

----- \(- 2\) +++++0++++++++++++

So from 1) we have \(x > 1\) so 2 intervals : one positive and one negative. two values. not suff

from2) \(x < - 2\) the only values are positive. so is suff.

B is the answer

Bunuel ?? correct my approach, pretty straightforward :)
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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Re: Is k^2 +k -2 > 0 ? [#permalink]

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New post 14 Jan 2015, 17:54
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Hi All,

If you spot the Quadratic expression in the prompt, then you can use an Algebra approach to get to the correct answer. This question can also be solved with a combination of TESTing VALUES and Number Properties.

We're asked if K^2 + K - 2 > 0. This is a YES/NO question

Fact 1: K < 1

IF....
K = 0
0^2 + 0 - 2 = -2 and the answer to the question is NO.

IF...
K = -3
(-3)^2 -3 - 2 = 4 and the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: K < -2

With this Fact, we have an interesting "limit" issue.

Even though it's not permitted....IF K = -2, then
(-2)^2 - 2 + 2 = 0 which is NOT > 0

As K becomes "more negative" (re. -2.1, -3, -100, etc.)....
K^2 creates a "bigger positive" than (+K - 2) creates a "negative"

eg.
K = -3
(-3)^2 = +9 vs. (-3 - 2)

K = -2.1
(-2.1)^2 = +4.41 vs. (-2.1 - 2)
Etc.

Thus, the result of the calculation will ALWAYS be greater than 0 and the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT.

Final Answer:
[Reveal] Spoiler:
B


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Re: Is k^2 +k -2 > 0 ? [#permalink]

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New post 19 May 2016, 04:52
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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New post 19 May 2016, 05:00
anartey wrote:
Is k^2 + k - 2 > 0 ?

(1) k < 1
(2) k < -2


For the equation k^2+k-2 greater than 0, K should either be less than -2 or should be greater than 1 (explaination: k^2+K-2 can be written as (k+1)(k-2) and for this product to be positive either K+1 and K-2 both should be positive or negative)

Coming to options:

1. K<1 is not sufficient because it includes the range k<-2 and some numbers beyond it
2. k<-2 is sufficient (as explained above)

Hope this helps! :)
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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New post 19 May 2016, 09:51
Statement 1 gives solutions both > or < 0

Statement 2 is enough to reach the solution.

B is the answer
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Re: Is k^2 +k -2 > 0 ?   [#permalink] 19 May 2016, 09:51
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